量子游荡作为经典随机游荡的量子类似物,在1993年由Aharonov等人提出^[1],它主要包括离散时间量子游荡和连续时间量子游荡.大量研究表明,量子游荡对量子算法的研究起非常重要的作用^[2-3].近几年来,量子游荡的渐进行为更是被广泛关注,见文献[4-6]. 2014年, Konno通过CGMV方法在文献[7]中得出空间齐次两态量子游荡的一致测度集包含于平稳测度集.同年, Endo等人利用SGF方法在文献[8]中得出了空间非齐次两态量子游荡的平稳测度及特征值问题. 2015年, Wang等人又利用SGF方法相继得出了空间非齐次单亏的三态Wojcik游荡的平稳测度,并发现在一些适当的条件下,它的平稳测度关于位置呈指数衰减^[9].不久, Endo等人在文献[10]中阐述了对角量子游荡的平稳测度.最近, Konno等人提出了一个新方法Reduced matrix,以此得出了离散时间三态量子游荡的一些性质^[11].本文将利用此方法进一步得出空间非齐次三态量子游荡的单相位和双相位模型的平稳测度及特征值问题.

在本章,我们介绍单相位三态量子游荡的单亏模型和双相位三态量子游荡的单亏模型及一些必要的定义.

定义2.1  空间非齐次三态量子游荡的单相位模型是定义在整数集${\Bbb Z}$上的,用一个手征空间$\{|L\rangle, |O\rangle, |R\rangle\}$和位置空间$\{|x\rangle:x\in{\Bbb Z}\}$来刻画.它的时间演化过程由下列$3\times3$阶酉矩阵决定

(2.1) $ \begin{equation}\label{1} U_{x}=\frac{{\rm e}^{{\rm i}\eta_{x}}}{3}\left( \begin{array}{ccc} -1&2&2 \\ 2 &~ -1~&2 \\ 2&2&-1 \\ \end{array} \right) , \quad x\in{\Bbb Z}, \end{equation}$

其中

(2.2) ${{\eta }_{x}}=\left\{ \begin{array}{*{35}{l}} 0,\quad x=\pm 1,\pm 2,\cdots , \\ 2\pi \tau ,~~x=0, \\\end{array} \right.$

且$\tau\epsilon(0, 1), \eta_{x}$表示游荡的相位为$2\pi\tau$.

设${\Bbb N}$是非负整数集, $\Psi_{n}(x)=(\Psi_{n}^{L}(x), \Psi_{n}^{O}(x), \Psi_{n}^{R}(x))^{T}$表示质点在$n\in{\Bbb N}$时刻,处于$x\in{\Bbb Z}$位置的波函数的振幅.下面我们讨论三态量子游荡的单相位单亏模型的平稳测度.显然,对每个位置$x\in{\Bbb Z}$, $U_{x} $都可以被分为三部分

$ U_{x}=U_{x}^{L}+U_{x}^{O}+U_{x}^{R}, $

其中

$ U_{x}^{L}=\frac{{\rm e}^{{\rm i}\eta_{x}}}{3}\left( \begin{array}{ccc} -1&~2~&2 \\ 0&0&0 \\ 0&0&0 \\ \end{array} \right) , ~~U_{x}^{O}=\frac{{\rm e}^{{\rm i}\eta_{x}}}{3}\left( \begin{array}{ccc} 0&0&0 \\ 2&~-1~&2 \\ 0&0&0 \\ \end{array} \right), ~~U_{x}^{R}=\frac{{\rm e}^{{\rm i}\eta_{x}}}{3}\left( \begin{array}{ccc} 0&0&0 \\ 0&~0~&0 \\ 2&2&-1 \\ \end{array} \right). $

通过这三个矩阵,我们可将波函数的时间演化过程定义为

(2.3) $\begin{equation}\Psi_{n+1}(x)=U_{x+1}^{L}\Psi_{n}(x+1)+U_{x}^{O}\Psi_{n}(x)+U_{x-1}^{R}\Psi_{n}(x-1), \end{equation}$

其中手征"$L$"和"$R$"分别表示游荡者向左和向右移动的手征, "$O$"表示游荡者移动到中性态的手征.设

$\Psi_{n}=(\cdots, \Psi_{n}(-1), \Psi_{n}(0), \Psi_{n}(1), \cdots)^{T}, ~~\Psi_{n}(x)=(\Psi_{n}^{L}(x), \Psi_{n}^{O}(x), \Psi_{n}^{R}(x))^{T}.$

$ U^{(s)}=\left( \begin{array}{ccccccc} \ddots&\vdots&\vdots&\vdots&\vdots&\vdots&\cdots \\ \cdots &~ U^{O}_{-2}~&U^{L}_{-1} &~ ~O~~&O&~~O~~&\cdots \\ \cdots&U^{R}_{-2}&U^{O}_{-1}&U^{L}_{0}&O&O&\cdots \\ \cdots&O&U^{R}_{-1}&U^{O}_{0}&U^{L}_{1}&O&\cdots \\ \cdots&O&O&U^{R}_{0}&U^{O}_{1}&U^{L}_{2}&\cdots \\ \cdots&O&O&O&U^{R}_{1}&U^{O}_{2}&\cdots \\ \cdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots \\ \end{array}, \right), ~~ O=\left( \begin{array}{ccc} 0 &~ 0~&0 \\ 0&0&0 \\ 0&0&0 \\ \end{array} \right), $

其中$T$表示转置,量子游荡在$n$时刻的态可表示为

(2.4) $\begin{equation}\Psi_{n}=(U^{(s)})^{n}\Psi_{0}, \quad n\geq0.\end{equation}$

下面定义平稳测度,先引入一个映射$\phi:({\Bbb C}^{3})^{{\Bbb Z}}\rightarrow{\Bbb R}_{+}^{{\Bbb Z}}$

$\phi(x)=(\cdots, |\Psi(-1)|^{2}, |\Psi(0)|^{2}, |\Psi(1)|^{2}, \cdots)^{T}\in{\Bbb R}_{+}^{{\Bbb Z}}, $

其中$\Psi=(\cdots, \Psi(-1), \Psi(0), \Psi(1), \cdots)^{T}\in({\Bbb C}^{3})^{{\Bbb Z}}$.对每个$x\in{\Bbb Z}$,我们有

$\phi(\Psi)(x)=|\Psi(x)|^{2}, \quad \Psi\in({\Bbb C}^{3})^{{\Bbb Z}}.$

显然,对固定的$\Psi\in({\Bbb C}^{3})^{{\Bbb Z}}$,函数$x\mapsto\phi(\Psi)(x)$通过$\mu(\cdot)=\phi(\Psi)(\cdot)$给出了$\mu$在${\Bbb Z}$上的一个测度.

对$n\geq0$,若定义$\mu_{n}(x)=\phi(\Psi_{n})(x), x\in{\Bbb Z}$,则$\mu_{n}$是${\Bbb Z}$上的一个测度,我们把这个概率测度便称为量子游荡在$n$时刻的测度,把$\mu_{n}(x)$看作是游荡者在$n$时刻,处于$x$位置的概率.

定义2.2  设${\cal M}_{s}=\{\phi(\Psi)\in{\Bbb R}_{+}^{{\Bbb Z}}|$存在$\Psi\in({\Bbb C}^{3})^{{\Bbb Z}}$,使得$\phi((U^{(s)})^{n}\Psi)=\phi(\Psi), n\geq0\}$.我们把${\cal M}_{s}$中的元素就称为量子游荡的平稳测度.

下面我们来考虑相应的特征值问题

(2.5) $U^{s}\Psi=\lambda\Psi(\lambda\in {\Bbb {\Bbb C}}), \quad n\geq0.$

显然,上式等价于下列问题

(2.6) $\lambda\Psi(x)=U_{x+1}^{L}\Psi(x+1)+U_{x}^{O}\Psi(x)+U_{x-1}^{R}\Psi(x-1) \quad (\lambda\in{\Bbb C}, |\lambda|=1).$

定义2.3  空间非齐次三态量子游荡的双相位模型是定义在整数集${\Bbb Z}$上的,用一个手征空间$\{|L\rangle, |O\rangle, |R\rangle\}$和位置空间$\{|x\rangle:x\in{\Bbb Z}\}$来刻画.它的时间演化由下面酉矩阵决定

(2.7) $U_{x}=\left\{\begin{array}{ll} U_+, \quad&x\geq1, \\ U_{0}, \quad &x=0, \\ U_-, \quad &x\leq-1, \end{array} \right.$

其中

(2.8) $U_{\pm}=\left( \begin{array}{ccc} -\frac{1+g_{\pm}}{2}&~~ \frac{\hbar\pm}{\sqrt{2}} ~~& \frac{1-g_{\pm}}{2} \\[3mm] \frac{\hbar\pm}{\sqrt{2}}&g_{\pm} & \frac{\hbar\pm}{\sqrt{2}} \\[3mm] \frac{1-g_{\pm}}{2} & \frac{\hbar\pm}{\sqrt{2}}&-\frac{1+g_{\pm}}{2} \end{array} \right), \quad U_{0}=\frac{\xi}{3}\left( \begin{array}{ccc} -1&2&2 \\ 2&~-1~&2 \\ 2&2&-1 \\ \end{array} \right). $

其中$g_{\pm}=\cos\gamma_{\pm}, \hbar_{\pm}=\sin\gamma_{\pm}, \gamma_{\pm}\in[0, 2\pi), \xi={\rm e}^{2\pi {\rm i} \tau}, \tau\in(0, 1).$

可以看出,由此酉矩阵所确定的量子coin的移位在正部和负部是不相同的,而且此量子游荡的行列式关于位置是独立的.我们便称此模型为"双相位三态量子游荡".

同单相位量子游荡一样,我们设

$\Psi_{n}=(\cdots, \Psi_{n}(-1), \Psi_{n}(0), \Psi_{n}(1), \cdots)^{T}, $

其中$\Psi_{n}(x)=(\Psi_{n}^{L}(x), \Psi_{n}^{O}(x), \Psi_{n}^{R}(x))^{T}$,

$ U^{(s)}=\left( \begin{array}{ccccccc} \ddots&\vdots&\vdots&\vdots&\vdots&\vdots&\cdots \\ \cdots &~ U^{O}_{-2} ~& U^{L}_{-1}&~~O~~&O&~~O~~&\cdots \\ \cdots&U^{R}_{-2}&U^{O}_{-1}&U^{L}_{0}&O&O&\cdots \\ \cdots&O&U^{R}_{-1}&U^{O}_{0}&U^{L}_{1}&O&\cdots \\ \cdots&O&O&U^{R}_{0}&U^{O}_{1}&U^{L}_{2}&\cdots \\ \cdots&O&O&O&U^{R}_{1}&U^{O}_{2}&\cdots \\ \cdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots \\ \end{array}, \right), ~~ O=\left( \begin{array}{ccc} 0&~0~&0 \\ 0&0&0 \\ 0&0&0 \\ \end{array} \right), $

其中$T$表示转置,在上式中

(2.9) $U_{x}^{L}=\left\{\begin{array}{ll} U_{x_{+}}^{L}, \quad& x\geq1 , \\ U_{x}^{L}, \quad& x=0 , \\ U_{x_{-}}^{L}, \quad& x\leq-1, \end{array} \right.~~ U_{x}^{O}=\left\{\begin{array}{ll} U_{x_{+}}^{O}, \quad &x\geq1, \\ U_{0}^{O}, \quad& x=0 , \\ U_{x_{-}}^{O}, \quad& x\leq-1, \end{array} \right.~~ U_{x}^{R}=\left\{\begin{array}{ll} U_{x_{+}}^{R}, \quad &x\geq1 , \\ U_{0}^{R}, \quad& x=0 , \\ U_{x_{-}}^{R}, \quad& x\leq-1, \end{array} \right.$

(2.10) $\begin{array}{l}U_{x_{+}}^{L}=\left( \begin{array}{ccc} -\frac{1+g_{+}}{2} &~~ \frac{\hbar+}{\sqrt{2}}~~&\frac{1-g_{+}}{2} \\ 0&0&0 \\ 0&0&0 \end{array} \right) , ~~ U_{x_{+}}^{O}=\left( \begin{array}{ccc} 0&0&0 \\\frac{\hbar+}{\sqrt{2}} &~~ g_{+}~~&\frac{\hbar+}{\sqrt{2}} \\ 0&0&0 \end{array} \right), \\ U_{x_{+}}^{R}=\left( \begin{array}{ccc} 0&0&0 \\ 0&0&0 \\\frac{1-g_{+}}{2} &~~ \frac{\hbar+}{\sqrt{2}}~~&-\frac{1+g_{+}}{2} \end{array} \right).\end{array}$

(2.11) $\begin{array}{l}U_{x_{-}}^{L}=\left( \begin{array}{ccc} -\frac{1+g_{-}}{2} &~~ \frac{\hbar-}{\sqrt{2}} ~~& \frac{1-g_{-}}{2} \\ 0&0&0 \\ 0&0&0 \end{array} \right) , ~~U_{x_{-}}^{O}=\left( \begin{array}{ccc} 0&0&0 \\\frac{\hbar-}{\sqrt{2}} &~~ g_{-} ~~& \frac{\hbar-}{\sqrt{2}} \\ 0&0&0 \end{array} \right), \\ U_{x_{-}}^{R}=\left( \begin{array}{ccc} 0&0&0 \\ 0&0&0 \\\frac{1-g_{-}}{2} &~~ \frac{\hbar-}{\sqrt{2}}~~&-\frac{1+g_{-}}{2} \end{array} \right).\end{array}$

(2.12) $U_{0}^{L}=\frac{\xi}{3}\left( \begin{array}{ccc} -1&2&2 \\ 0&~0~&0 \\ 0&0&0 \\ \end{array} \right) , ~~U_{0}^{O}=\frac{\xi}{3}\left( \begin{array}{ccc} 0&0&0 \\ 2&~-1~&2 \\ 0&0&0 \\ \end{array} \right), ~~U_{0}^{R}=\frac{\xi}{3}\left( \begin{array}{ccc} 0&0&0 \\ 0&~0~&0 \\ 2&2&-1 \\ \end{array} \right).$

其中初始量子态为

$\Psi_{0}=(\cdots, \Psi_{0}(-1), \Psi_{0}(0), \Psi_{0}(1), \cdots)^{T}, $

其中$\Psi_{0}(x)=(\Psi_{0}^{L}(x), \Psi_{0}^{O}(x), \Psi_{0}^{R}(x))^{T}.$

相应的特征值问题等价于

(2.13) $\lambda\Psi(x)=U_{x+1}^{L}\Psi(x+1)+U_{x}^{O}\Psi(x)+U_{x-1}^{R}\Psi(x-1)\quad (\lambda\in{\Bbb C}, |\lambda|=1).$

对于一个定义在整数集${\Bbb Z}$上离散时间的三态量子游荡,它的时间演化过程由酉矩阵A所确定

$ A=\left( \begin{array}{ccc} a_{11}&~~a_{12}~~&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \\ \end{array} \right).$

设

$ B=\left|\begin{array}{cc} a_{11} &~~ a_{12} \\ a_{21}&~~a_{22} \end{array}\right|, ~~ C=\left|\begin{array}{cc} a_{12}&~~a_{13} \\ a_{22}&~~a_{23} \end{array}\right|, ~~D=\left|\begin{array}{cc} a_{21}&~~a_{22} \\ a_{31}&~~a_{32} \end{array}\right|, ~~ E=\left|\begin{array}{cc} a_{22}&~~a_{23} \\ a_{32}&~~a_{33} \end{array}\right|, $

且$a_{ij}\neq0\ (i, j=1, 2, 3).$

引理3.1^[11]  假设$\lambda=-\frac{C}{a_{13}}=-\frac{D}{a_{31}}$, $\lambda$为酉矩阵A对应的特征值且$|\lambda|=1, $初始态$\Psi^{L}_{0}=\varphi_{1}, \Psi^{R}_{0}=\varphi_{3}$,则

(3.1) $\Psi(x)=\left[\begin{array}{c} (\tilde{a_{1}}^{-1}\lambda)^{x}\varphi_{1} \\ -\frac{a_{13}}{a_{12}a_{23}}\{a_{21}(\tilde{a_{1}}^{-1}\lambda)^{x}\varphi_{1}+a_{23}(\tilde{a_{2}}\lambda^{-1})^{x}\varphi_{3}\} \\ (\tilde{a_{2}}\lambda^{-1})^{x}\varphi_{3} \end{array} \right], $

其中

$\begin{eqnarray*} \tilde{a_{1}}=a_{11}-\frac{a_{13}a_{21}}{a_{23}}, \quad \tilde{a}_{2}=a_{33}-\frac{a_{23}a_{31}}{a_{21}}.\end{eqnarray*} $

引理3.2^[11]  假设$\lambda=\frac{B}{a_{11}}=\frac{E}{a_{33}}$, $|\lambda|=1$且$\lambda^{2}=\tilde{a}_{1}\tilde{a}_{2}$,令$\{\varphi_{x}\}_{x\in{\Bbb Z}}$为一列复数序列, $\Psi^{L}(x)=\varphi_{x}\in C(x\in Z), $不包括$\varphi\equiv {0}$,则

(3.2) $ \Psi(x)=\left[\begin{array}{c} \varphi_{x} \\ -\frac{a_{11}}{a_{12}a_{21}}\{a_{21}\varphi_{x}+a_{23}(\tilde{a_{1}}^{-1}\lambda)\varphi_{x-1}\} \\ (\tilde{a_{1}}^{-1}\lambda)\varphi_{x-1} \end{array} \right] \quad (x\in{\Bbb Z}), $

其中

$ \tilde{a_{1}}=a_{13}-\frac{a_{11}a_{23}}{a_{21}}, \quad \tilde{a}_{2}=a_{31}-\frac{a_{21}a_{33}}{a_{23}}. $

接下来我们考虑单相位单亏模型的平稳测度.

定理3.3  令$\Psi_{n}(x)=(\Psi_{n}^{L}(x), \Psi_{n}^{O}(x), \Psi_{n}^{R}(x))^{T}$为波函数的概率振幅,初始态$\Psi^{L}_{0}=\varphi_{1}, \Psi^{R}_{0}=\varphi_{3}, \varphi_{1}, \varphi_{3}\in{\Bbb C}^{2}$且$|\varphi_{1}|^{2}+|\varphi_{3}|^{2}>0$.在定义$2.1$给出的模型中

$ U_{x}=\frac{1}{3}\left( \begin{array}{ccc} -1&2&2 \\ 2 &~~ -1~~&2 \\ 2&2&-1 \\ \end{array} \right), \quad x\neq0;$

$ U_{0}=\frac{\xi}{3}\left( \begin{array}{ccc} -1&2&2 \\ 2 &~~ -1~~&2 \\ 2&2&-1 \\ \end{array} \right), \quad x=0, $

其中$\xi={\rm e}^{2\pi{\rm i}\tau}, \tau\in(0, 1)$.则平稳测度

$ \mu(x)=2\{|\varphi_{1}|^{2}+|\varphi_{3}|^{2}+{\cal R}e(\Delta\varphi_{1}\bar{\varphi_{3}})\}, $

其中$\Delta=\bar{\xi}, x=-1; \quad \Delta=\xi, x=1;\quad \Delta=1$,其它.

可以看出此平稳测度为非一致测度.

证  由第二章内容可知,解方程(2.5)等价于解方程(2.6),即

(3.3) $\lambda\Psi(x)=U_{x+1}^{L}\Psi(x+1)+U_{x}^{O}\Psi(x)+U_{x-1}^{R}\Psi(x-1) \quad (\lambda\in{\Bbb C}, |\lambda|=1).$

其中$x\neq0$时,有

(3.4) $ U_{x}^{L}=\frac{1}{3}\left( \begin{array}{ccc} -1&2&2 \\ 0&~~0 ~~& 0 \\ 0&0&0 \\ \end{array} \right) , ~~U_{x}^{O}=\frac{1}{3}\left( \begin{array}{ccc} 0&0&0 \\ 2 &~~ -1~~&2 \\ 0&0&0 \\ \end{array} \right), ~~U_{x}^{R}=\frac{1}{3}\left( \begin{array}{ccc} 0&0&0 \\ 0 &~~ 0~~&0 \\ 2&2&-1 \\ \end{array} \right). $

$x=0$时,有

(3.5) $ U_{0}^{L}=\frac{\xi}{3}\left( \begin{array}{ccc} -1 &~~ 2~~&2 \\ 0&0&0 \\ 0&0&0 \\ \end{array} \right) , ~~U_{0}^{O}=\frac{\xi}{3}\left( \begin{array}{ccc} 0&0&0 \\ 2 &~~ -1~~&2 \\ 0&0&0 \\ \end{array} \right), ~~U_{0}^{R}=\frac{\xi}{3}\left( \begin{array}{ccc} 0&~~0~~&0 \\ 0&0&0 \\ 2&2&-1 \\ \end{array} \right). $

结合(3.3)-(3.5)式可知:

当$x\neq0, \pm1$时,有

$ \lambda\Psi^{L}(x)=-\frac{1}{3}\Psi^{L}(x+1)+\frac{2}{3}\Psi^{O}(x+1)+\frac{2}{3}\Psi^{R}(x+1), $

$\lambda\Psi^{O}(x)=\frac{2}{3}\Psi^{L}(x)-\frac{1}{3}\Psi^{O}(x)+\frac{2}{3}\Psi^{R}(x), $

$\lambda\Psi^{R}(x)=\frac{2}{3}\Psi^{L}(x-1)+\frac{2}{3}\Psi^{O}(x-1)-\frac{1}{3}\Psi^{R}(x-1).$

由引理$3.1$可以得$\lambda=-1, \tilde{a}_{1}=\tilde{a}_{2}=-1$,则相应的简化矩阵为

$ A_{x}=\left( \begin{array}{cc} -1 &~~ 0 \\ 0 &~~ -1 \\ \end{array} \right).$

则

$ \Psi(x)=\left[\begin{array}{c} \varphi_{1} \\ -(\varphi_{1}+\varphi_{3}) \\ \varphi_{3} \end{array} \right], \quad \mu(x)=2\{|\varphi_{1}|^{2}+|\varphi_{2}|^{2}+{\cal R}e(\varphi_{1}\bar{\varphi_{3}})\}.$

同理, $x=1$时,可得$\lambda=-1, \tilde{a}_{1}=-1, \tilde{a}_{2}=-\xi.$则相应的简化矩阵为

$ A_{1}=\left( \begin{array}{cc} -1 &~~ 0 \\ 0 &~~ -\xi \\ \end{array} \right).$

故

$\Psi(1)=\left[\begin{array}{c} \varphi_{1} \\ -(\varphi_{1}+\xi\varphi_{3}) \\ \xi\varphi_{3} \end{array} \right], \quad \mu(1)=2\{|\varphi_{1}|^{2}+|\varphi_{3}|^{2}+{\cal R}e(\bar{\xi}\varphi_{1}\bar{\varphi_{3}})\}.$

$x=-1$时,可得$\lambda=-1, \tilde{a}_{1}=-\xi, \tilde{a}_{2}=-1.$则相应的简化矩阵为

$ A_{-1}=\left( \begin{array}{cc} -\xi &~~ 0 \\ 0 &~~ -1 \\ \end{array} \right).$

故

$ \Psi(-1)=\left[\begin{array}{c} \xi\varphi_{1} \\ -(\xi\varphi_{1}+\varphi_{3}) \\ \varphi_{3} \end{array} \right], \quad \mu(-1)=2\{|\varphi_{1}|^{2}+|\varphi_{3}|^{2}+{\cal R}e(\xi\varphi_{1}\bar{\varphi_{3}})\}.$

$x=0$时,可得$\lambda=-\xi, \tilde{a}_{1}=\tilde{a}_{2}=-1.$则相应的简化矩阵为

$ A_{0}=\left( \begin{array}{cc} -1 &~~ 0 \\ 0 &~~ -1 \\ \end{array} \right).$

故

$ \Psi(0)=\left[\begin{array}{c} \varphi_{1} \\ -(\varphi_{1}+\varphi_{3}) \\ \varphi_{3} \end{array} \right], \quad \mu(0)=2\{|\varphi_{1}|^{2}+|\varphi_{3}|^{2}+{\cal R}e(\varphi_{1}\bar{\varphi_{3}})\}.$

证毕.

下面我们考虑用引理$3.2$来讨论单相位模型的平稳测度:当$x\neq0, \pm1$时,由引理$3.2$可以得$\lambda=1, \tilde{a}_{1}=\tilde{a}_{2}=1$,则相应的简化矩阵为

$ A_{x}=\left( \begin{array}{cc} 0 &~~ 1 \\ 1 &~~ 0 \\ \end{array} \right), $

则

$ \Psi(x)=\left[\begin{array}{c} \varphi_{x} \\ \frac{1}{2}(\varphi_{x}+\varphi_{x-1}) \\ \varphi_{x-1} \end{array} \right], \quad \mu(x)=\frac{5}{4}\{|\varphi_{x}|^{2}+|\varphi_{x-1}|^{2}\}+\frac{1}{2}\{{\cal R}e(\varphi_{x-1}\bar{\varphi_{x}})\}. $

当$x=1$时, $\lambda=1, \tilde{a}_{1}=1, \tilde{a}_{2}=\xi, $由于$\lambda^{2}\neq\tilde{a}_{1}\tilde{a}_{2}$,故不满足条件;同理, $x=0, -1$时也不满足.故引理$3.2$不适用此模型.

下面讨论两相位单亏模型的平稳测度.

定理3.4  设$\Psi_{n}(x)=(\Psi_{n}^{L}(x), \Psi_{n}^{O}(x), \Psi_{n}^{R}(x))^{T}$为两相位三态量子游荡的波函数的概率振幅,令初始态$\Psi^{L}_{0}=\varphi_{1}, \Psi^{R}_{0}=\varphi_{3}, \varphi_{1}, \varphi_{3}\in{\Bbb C}^{2}$且$|\varphi_{1}|^{2}+|\varphi_{3}|^{2}>0$.对定义$2.3$给出的双相位单亏模型,可以得出平稳测度

$\mu(x)=\Delta_{1}\{|\varphi_{1}|^{2}+|\varphi_{3}|^{2}\}+\Delta_{2}{\cal R}e(\Delta_{3}\varphi_{1}\bar{\varphi_{3}})\}, $

其中

$ \left\{\begin{array}{ll}\Delta_{1}=2, \Delta_{2}=2, \Delta_{3}=\xi, \quad &x=-1; \\\Delta_{1}=\frac{(g_{+}-1)^{2}}{2\hbar_{+}^{2}}+1, \Delta_{2}=\frac{(g_{+}-1)^{2}}{\hbar_{+}^{2}}, \Delta_{3}=\bar{\xi}, \quad &x=1; \\[3mm]\Delta_{1}=\frac{(g_{\pm}-1)^{2}}{2\hbar_{\pm}^{2}}+1, \Delta_{2}=\frac{(g_{\pm}-1)^{2}}{\hbar_{\pm}^{2}}, \Delta_{3}=1, \quad &x\neq\pm1. \end{array} \right.$

证  结合(2.9)-(2.13)式可知

当$x\neq0, \pm1$时,有

$\lambda\Psi^{L}(x)=-\frac{1+g_{\pm}}{2}\Psi^{L}(x+1)+\frac{\hbar_{\pm}}{\sqrt{2}}\Psi^{O}(x+1)+\frac{1-g_{\pm}}{2}\Psi^{R}(x+1), $

$\lambda\Psi^{O}(x)=\frac{\hbar_{\pm}}{\sqrt{2}}\Psi^{L}(x)+g_{\pm}\Psi^{O}(x)+\frac{\hbar_{\pm}}{\sqrt{2}}\Psi^{R}(x), $

$\lambda\Psi^{R}(x)=\frac{1-g_{\pm}}{2}\Psi^{L}(x-1)+\frac{\hbar_{\pm}}{\sqrt{2}}\Psi^{O}(x-1)-\frac{1+g_{\pm}}{2}\Psi^{R}(x-1).$

由引理$3.1$可以得$\lambda=-1, \tilde{a}_{1}=\tilde{a}_{2}=-1$,则相应的简化矩阵为

$ B_{x}=\left( \begin{array}{cc} -1~~ &~~ 0 \\ 0 &-1 \\ \end{array} \right).$

则

$ \Psi(x)=\left[\begin{array}{c} \varphi_{1} \\ \frac{(g_{\pm}-1)}{\sqrt{2}\hbar_{\pm}}(\varphi_{1}+\varphi_{3}) \\ \varphi_{3} \end{array} \right], $

$\mu(x)=\bigg\{\frac{(g_{\pm}-1)^{2}}{2\hbar_{\pm}^{2}}+1\bigg\}\{|\varphi_{1}|^{2}+|\varphi_{3}|^{2}\}+\frac{(g_{\pm}-1)^{2}}{\hbar_{\pm}^{2}}{\cal R}e(\varphi_{1}\bar{\varphi_{3}}).$

同理, $x=1$时,可得$\lambda=-1, \tilde{a}_{1}=-1, \tilde{a}_{2}=-\xi.$则相应的简化矩阵为

$ B_{1}=\left( \begin{array}{cc} -1~~ &~~ 0 \\ 0 &-\xi \\ \end{array} \right).$

故

$ \Psi(1)=\left[\begin{array}{c} \varphi_{1} \\ \frac{(g_{+}-1)}{\sqrt{2}\hbar_{+}}(\varphi_{1}+\xi\varphi_{3}) \\ \xi\varphi_{3} \end{array} \right], $

$\mu(1)=\bigg\{\frac{(g_{+}-1)^{2}}{2\hbar_{+}^{2}}+1\bigg\}\{|\varphi_{1}|^{2}+|\varphi_{3}|^{2}\}+\frac{(g_{+}-1)^{2}}{\hbar_{+}^{2}}{\cal R}e(\bar{\xi}\varphi_{1}\bar{\varphi_{3}}).$

$x=-1$时,可得$\lambda=-1, \tilde{a}_{1}=-\xi, \tilde{a}_{2}=-1.$则相应的简化矩阵为

$ B_{-1}=\left( \begin{array}{cc} -\xi ~~&~~ 0 \\ 0 &-1 \\ \end{array} \right).$

故

$ \Psi(-1)=\left[\begin{array}{c} \xi\varphi_{1} \\ -(\xi\varphi_{1}+\varphi_{3}) \\ \varphi_{3} \end{array} \right], ~~\mu(-1)=2\{|\varphi_{1}|^{2}+|\varphi_{3}|^{2}+{\cal R}e(\xi\varphi_{1}\bar{\varphi_{3}})\}. $

$x=0$时,可得$\lambda=\frac{2\sqrt{2}}{3}\frac{\hbar_{+}\xi}{1-g_{+}}+\frac{\xi}{3}, \tilde{a}_{1}=\tilde{a}_{2}=-1.$则相应的简化矩阵为

$ B_{0}=\left( \begin{array}{cc} -1 ~~&~~ 0 \\ 0 &-1 \\ \end{array} \right). $

故

$ \Psi(0)=\left[\begin{array}{c} \varphi_{1} \\ \frac{g_{+}-1}{\sqrt{2}\hbar_{+}}(\varphi_{1}+\varphi_{3}) \\ \varphi_{3} \end{array} \right], $

$\mu(0)=\bigg\{\frac{(g_{+}-1)^{2}}{2\hbar_{+}^{2}}+1\bigg\}\{|\varphi_{1}|^{2}+|\varphi_{3}|^{2}\}+\frac{(g_{+}-1)^{2}}{\hbar_{+}^{2}}{\cal R}e(\varphi_{1}\bar{\varphi_{3}}).$

证毕.

与单相位模型的类似,引理$3.2$的结论也不适用于双相位模型.